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I'm given a zero-pole plot (Real/Imaginary Plane) of a system in which poles occur at $\frac{-9}{10}$, $\frac{9}{10}$, and zeros occur at $\frac{10}{9}j$, $\frac{-10}{9}j$. An additional zero occurs at negative infinity.

I'm also given that $H_{F}(f)=1$ for $f=0$, where $H_{F}(f)$ refers to $H(z)$ for when $z=e^{-j 2\pi f}$. I'm tasked to find $H(z)$.

I know I can express $H(z)$ as $$H(z)=G_0z^{-1}\frac{(z-z_0)(z-z_1)\ldots}{(z-p_0)(z-p_1)\ldots} $$ for zeros $$z_0,z_1,\ldots$$ and poles $$p_0, p_1,\ldots$$

I can find the constant $G_0$ by using the fact $H_{F}(f)=1$ for $f=0$.

I'm unsure how to express the fact that the system has a zero at negative infinity.

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  • $\begingroup$ Why do you specify that the zero is at negative infinity? A zero at infinity would mean that $H(z) \xrightarrow{z\to\infty}0$. This is the same as saying that the system has a pole at the origin. $\endgroup$
    – Tendero
    Commented Apr 5, 2016 at 0:05
  • $\begingroup$ Yes, correct SleuthEye. Thanks for catching that. $\endgroup$
    – Minh Tran
    Commented Apr 5, 2016 at 1:01
  • $\begingroup$ M. S. That was given in the prompt for the problem. I just e-mailed my Prof. and he said that if we were to represent $H(z)$ as $\frac{\sum_{m=0}^{M} b_m z^{-m}}{1+\sum_{k=1}^{N} a_k z^{-k}}$ = $\frac{b_0+b_1 z^{-1}+\dots +b_M z^{-M}}{1+a_1 z^{-1}+\dots+a_N z^{-N}}$, a zero at negative infinity means that $b_0=0$. $\endgroup$
    – Minh Tran
    Commented Apr 5, 2016 at 1:10
  • $\begingroup$ We talk about "negative infinity" when calculating limits, for example, in real calculus. When dealing with complex numbers, the "point at infinity" is unique and is neither positive or negative. If you want more information about this topic, you can investigate about the extended complex plane. Regarding the last equation you wrote, the fact that that $H(z)$ has or not a zero at infinity depends on the values of $M$ and $N$. Only if the denominator is of higher order than the numerator will there be zero(s) at infinity. $\endgroup$
    – Tendero
    Commented Apr 5, 2016 at 15:11

2 Answers 2

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The $z^{-1}$ term in the $H(z)$ expression you provided is the expression of the zero at infinity:

$$\lim_{z\to\infty} \frac{1}{z} = 0$$

And in your specific case:

$$\lim_{z\to\infty} H(z) = G_{0}\frac{1}{z}\frac{(z-\frac{10}{9}j)(z+\frac{10}{9}j)}{(z-\frac{9}{10})(z+\frac{9}{10})} = 0$$

Generally, if the order of the denominator exceeds the order of the numerator, then you have 1 or more zeros at $\infty$

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First, express the zeros and poles in a polar form: $z=re^{-j 2\pi f}$ You can do this using the Euler formula

Second, substitute $z=e^{-j 2\pi f}$ in $H(z)$, including the values for the poles and zeros. You will have $H(f)$ now. Everything should be expressed as a function of $f$ and $G_0$ at this point.

Third, substitute $f=0$ in $H(f)$ and do $H(f)=1$. Then you will be able to get the value for $G_0$.

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